Zodiackillersite.com

doranchak wrote:OK, I cracked it (daikon, please confirm the solution I pm'd you), but it required many manual steps. Hint: Reduce the multiplicity before trying to crack this.

Yes, congrats, you did solve it! Well done. It's not the cleanest solution, but close enough to read the information I was trying to convey in it. I'm quite amazed! I was actually convinced that it will never be cracked as I had to reduce the unique symbol count down so low, before ZKD/AZD were able to crack it automatically, that it was mostly a monoalphabetic substitution at that point! I'm guessing you merged repeated cycles manually to improve multiplicty, right? How many unique symbols did you end up with before it was cracked?

I got it down to 49 by identifying two really strong cycles and merging their constituent symbols. It was enough to get one of the solvers to produce a few small bits and pieces of recognizable text. I then saw what you were going for in the plain text and did a manual solve for the rest.

I identified four cycles in total but was afraid of false positives.

I think if this cipher did not have the cycles in it, the solution would be out of reach (hopefully only temporarily). You should encode another one with the cycles randomized.

Jarlve wrote:I think you tried to induce high multiplicity by assigning some high counts to a few symbols and low counts to the bulk.

That's not quite it, but you are on the right track. The classical way of creating homophonic substitution ciphers is to assign many ciphertext symbols to high frequency letters in the plaintext so that it has the "smoothing" effect on the frequency distribution in the ciphertext. Well, Z340 has one symbol (+) with peculiarly high frequency. We still don't know why, so I can't draw any real conclusions from it, but I thought I'll try to do the same thing in my cipher to just mimic the same behavior.

One way of doing it is to follow the classic method of "more ciphertext symbols for higher frequency letters", save for one somewhat frequent letter, to which we will assign just one ciphertext symbol. We obviously don't want it to be "e" as it would be an easy guess, but the rest of frequent letters in English texts have roughly equal distribution, so we can pick one and still have enough leeway in confusing the decryptors. It turned out to be quite a stumbling block for hill-climb based auto-solvers too, but for a different reason I believe, as hill-climb algorithm (HCA) doesn't care about frequencies in any way. I think it has to do with smoothness of the solution field. Basically, HCA expects that a small change in the inputs (i.e. changing one of the decryption key symbols) would result in a relatively small change in the fitness/score of the solution. It won't be true any more in case we do assign just one ciphertext symbol to a high frequency letter. Because changing that symbol in the key would result in a huge change in the fitness/score of the solution, as it will affect/change many parts of the ciphertext, being repeated so many times in the ciphertext. So HCA will often jump over the correct solution without realizing it, will backtrack, and will never approach the real top of the hill. It will eventually find the top, I believe, but it might take a lot of tries, close to brute force iterating through all possible keys.

Now, I have no idea if that's what Z was doing with the "+", or how did he know about HCA back in the '60s, but it's possible that he just stumbled on something that's hard to crack using HCA by pure chance. Or it could be that Z used "+" for something entirely different, or that it can't be cracked because it's not a straight substitution cipher.

glurk wrote:So, in this one, the + symbol must be either E, T, A, O, I, or N.

It's the last one, I didn't want to make it too easy. :)

glurk wrote:Just to add, this is honestly the strangest cipher I've ever seen. If it turns out to be fairly normal English text, I bow down to you.

Well, that was my thinking process. In what way can plaintext be "special", but still be an English text and something that Z could possibly write? I'll give you a hint, I took inspiration from Z32 "map cipher" as he hinted that it "concerns radians & inches along the radians".

Jarlve wrote:Maybe the repeats/patterns can be explained by the plaintext being very repetitive. The same thing repeating over and over again, or use of a very limited vocabulary.

You've hit the head on the nail! :) It's not the same thing repeated over and over, but the vocabulary is very limited indeed. You'd think that it would make it easy to crack automatically, but it seems to be the other way around.

doranchak wrote:I think if this cipher did not have the cycles in it, the solution would be out of reach (hopefully only temporarily). You should encode another one with the cycles randomized. :)

Yes, you are correct, randomizing cycles would definitely help. But it would only make the manual solving more difficult. Auto-solvers don't use cycles in any way, as best as I can tell. So it would have no effect for them. And the whole point of this exercise was to show that there are cases of homophonic substitution ciphers that can't be cracked automatically. Not by current hill-climb based algorithms anyway. Whether Z340 is such a special case of a straight homophonic substitution cipher, or it uses some extra steps to obfuscate plaintext, it still remains to be seen. I just wanted to hopefully spur further improvements to the automatic solvers, as the current prevailing misconception seems to be that they are already powerful enough to solve any homophonic substitution cipher.

I've got this now, I guess I was close earlier after seemingly finding EIGHT, HUNDRED, FIFTY, etc... I should have stuck with it.

I didn't know if I was on the right track or not, after a while your eyes start to glaze over, LOL.

Daikon, can you PM me the exact plaintext? Trying to get this all correct as you intended is horrible! I want to try an experiment or two using the exact solution.

-glurk

PS. I like the bookends.

EDIT: I actually think I have it exact now. I've sent you a PM.

Well done doranchak and glurk!

daikon,

As I've said before I believe the limited vocabulary artificially raise the multiplicity. So it's a multiplicity problem more than anything else. I agree that some ciphers may be unsolveable due to the nature of the plaintext and multiplicity of the cipher. It's fair to assume that such a scheme is not actual in the 340 because there are not much patterns and it scores much lower than your 3rd cipher in both of the solvers.

To answer your question from the other thread, I'm planning to release the next version of my solver near the end of the month. It will have 3 modules for now, the old solver, a new 4-gram solver and a 5-gram solver. I'm spending most of my time optimizing the 5-gram solver to improve recovery rate for various ciphers, yours included.

Edit: and btw, I'm really thankful for your ciphers and work on this. It's something I needed.

glurk wrote:EDIT: I actually think I have it exact now. I've sent you a PM.

You do have it *exact* to the letter. Wow! I had to abbreviate degrees/minutes/seconds to just one letter, as the message was already getting too long and I wanted to fit everything in. I figured one single letter every few words shouldn't complicate things too much for auto-solvers, and Zodiac did shorten his words on multiple occasions.

I think no need to post the plaintext here any more? In case someone else comes across it later on and wants to solve it independently. I've given so many "clews" in this thread anyway, it's practically solves itself now. :)

glurk wrote:I've got this now, I guess I was close earlier after seemingly finding EIGHT, HUNDRED, FIFTY, etc... I should have stuck with it.

So, how did you solve it, I'm curious? Similar to doranchak, by merging repeating cycles, i.e. guessing homophones that solve to the same plaintext letter?